Give two different sequences of three transformations that would map PQR onto EFG given that PQR=EFG.

Mathematics

2 answers:

Gre4nikov [31]2 years ago

3 0

9514 1404 393

Answer:

Translate P to E; rotate ∆PQR about E until Q is coincident with F; reflect ∆PQR across EF
Reflect ∆PQR across line PR; translate R to G; rotate ∆PQR about G until P is coincident with E

Step-by-step explanation:

The orientations of the triangles are opposite, so a reflection is involved. The various segments are not at right angles to each other, so a rotation other than some multiple of 90° is involved. A translation is needed in order to align the vertices on top of one another.

The rotation is more easily defined if one of the ∆PQR vertices is already on top of its corresponding ∆EFG vertex, so that translation should precede the rotation. The reflection can come anywhere in the sequence.

<em>Additional comment</em>

The mapping can be done in two transformations: translate a ∆PQR vertex to its corresponding ∆EFG point; reflect across the line that bisects the angle made at that vertex by corresponding sides.

WARRIOR [948]2 years ago

3 0

Answer:

Translate P to E; rotate ∆PQR about E until Q is coincident with F; reflect ∆PQR across EF

Reflect ∆PQR across line PR; translate R to G; rotate ∆PQR about G until P is coincident with E

Step-by-step explanation:

Additional comment

The mapping can be done in two transformations: translate a ∆PQR vertex to its corresponding ∆EFG point; reflect across the line that bisects the angle made at that vertex by corresponding sides.

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A standard deck of cards contains 52 cards. One card is selected from the deck.(a)Compute the probability of randomly selecting

Aneli [31]

a. There are four 5s that can be drawn, and $\binom43=4$ ways of drawing any three of them. There are $\binom{52}3=22,100$ ways of drawing any three cards from the deck. So the probability of drawing three 5s is

$\dfrac{\binom43}{\binom{52}3}=\dfrac4{22,100}=\dfrac1{5525}\approx0.00018$

In case you're asked about the probability of drawing a 3 or a 5 (and NOT three 5s), then there are 8 possible cards (four each of 3 and 5) that interest you, with a probability of $\frac8{52}=\frac2{13}\approx0.15$ of getting drawn.

b. Similar to the second case considered in part (a), there are now 12 cards of interest with a probability $\frac{12}{52}=\frac3{13}\approx0.23$ of being drawn.

c. There are four 6s in the deck, and thirteen diamonds, one of which is a 6. That makes 4 + 13 - 1 = 16 cards of interest (subtract 1 because the 6 of diamonds is being double counted by the 4 and 13), hence a probability of $\frac{16}{52}=\frac4{13}\approx0.31$ .